http://acm.hdu.edu.cn/showproblem.php?pid=1455
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=243
uva开头描述：
307 - Sticks
Time limit: 3.000 seconds
hduoj 开头描述：
E - Sticks
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
题目描述：
Description
George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.
Input
The input file contains blocks of 2 lines. The first line contains the number of sticks parts after cutting. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.
Output
The output file contains the smallest possible length of original sticks, one per line.
Sample Input
9 5 2 1 5 2 1 5 2 1 4 1 2 3 4 0
Sample Output
6 5 分析：uva会TLE ， hduoj AC
AC代码：
1 #include<algorithm>
2 #include<cstring>
3 #include<cstdio>
4 #include<iostream>
5 using namespace std;
6
7 const int N = 1e4+10;
8 int len[N],sum,L,T;
9 int used[N];
10
11 int cmp(const void *a,const void *b)
12 {
13 return *(int *)b-*(int *)a;
14 }
15
16 bool DFS(int m,int left)
17 //m为剩余的木棒数，left为当前正在拼接的木棒和假定的木棒长度L还缺少的长度
18 {
19 if(m == 0 && left == 0)
20 return true;
21 if(left == 0)//一根刚刚拼完
22 left = L;
23 for(int i=0; i<T; i++)
24 {
25 if(!used[i] && len[i]<=left)
26 {
27 if(i>0)//如果前者已经出现过的不能用，则当前的也不能用
28 {
29 if(!used[i-1] && len[i] == len[i-1])
30 continue;
31 }
32 used[i] = 1;
33 if(DFS(m-1,left-len[i]))
34 return true;
35 else
36 {
37 used[i] = 0;
38 if(len[i] == left || left == L)
39 return false;
40 }
41 }
42 }
43 return false;
44 }
45
46 int main()
47 {
48 while(scanf("%d",&T) && T)
49 { sum = 0 ;
50 for(int i=0;i<T;i++)
51 {
52 scanf("%d",&len[i]);
53 sum = sum + len[i];
54 }
55 //sort(len,len+T,cmp); sort 超时
56 qsort(len,T,sizeof(int),cmp); //从大到小排序
57 for(L = len[0];L<=sum/2;L++)
58 {
59 if(sum%L)
60 continue;
61 memset(used,0,sizeof(used));
62 if(DFS(T,L))
63 {
64 printf("%d\n",L);
65 break;
66 }
67 }
68 if(L>sum/2)
69 printf("%d\n",sum);
70 }
71 return 0;
72 }